Theorem 1.

For ⇐, we will need a lemma that essentially says that a submodule of M is uniquely determined by its image in M′′ and its intersection with M′:

Lemma 1.

In the situation of the theorem, if N1,N2⊂M are submodules with N1⊂N2, π⁢(N1)=π⁢(N2), and N1∩ι⁢(M′)=N2∩ι⁢(M′), then N1=N2.

Proof.

The proof is essentially a diagram chase.
Choose x∈N2. Then π⁢(x)=π⁢(x′) for some x′∈N1, and thus π⁢(x-x′)=0, so that x-x′∈im⁡ι, and x-x′∈N2 since N1⊂N2. Hence x-x′∈N2∩ι⁢(M′)=N1∩ι⁢(M′)⊂N1. Since x′∈N1, it follows that x∈N1 so that N1=N2.
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Proof.

(⇒): If M is Noetherian (Artinian), then any ascending (descending) chain of submodules of M′ (or of M′′) gives rise to a similar sequence in M, which must therefore terminate. So the original chain terminates as well.
(⇐): Assume first that M′,M′′ are Noetherian, and choose any ascending chain M1⊂M2⊂… of submodules of M. Then the ascending chain π⁢(M1)⊂π⁢(M2)⊂… and the ascending chain M1∩ι⁢(M′)⊂M2∩ι⁢(M′)⊂… both stabilize since M′ and M′′ are Noetherian. We can choose n large enough so that both chains stabilize at n. Then for N≥n, we have (by the lemma) that MN=Mn since π⁢(MN)=π⁢(Mn) and MN∩ι⁢(M′)=Mn∩ι⁢(M′). Thus M is Noetherian. For the case where M is Artinian, an identical proof applies, replacing ascending chains by descending chains.
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